# Thread: Probability Thread

1. Originally Posted by ben1996123
Probably only Ryan Heise knows. E-mail him here: webmaster at hi-games.net

2. Originally Posted by Johan444
Probably only Ryan Heise knows. E-mail him here: webmaster at hi-games.net
Why? Unless I've misunderstood the question, just divide the number of 2-move QTM scrambles (I think it's 27) by the number of combinations?

3. What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)

4. I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?

5. Originally Posted by ben1996123
1/408 240

6. Originally Posted by Marcell
I'm not sure I get this - sorry if it's just me. Anyway: By stickers do you mean edge + corner stickers as well? Why is x between 0 and 10? What do you mean ignoring twisted corners - we leave them there twisted? And what about flipped edges?
I mean only corners. And yes, completely ignoring twisted corners as if they are solved.

x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10

7. Right, thanks. Now let me think about it

8. Originally Posted by riffz
What is the probability in a BLD solve that, ignoring twisted corners in place and using a fixed buffer, you will have to shoot to x stickers? (0 <= x <= 10)
Originally Posted by riffz
I mean only corners. And yes, completely ignoring twisted corners as if they are solved.

x is 10 or less because that is the maximum number of stickers you would have to shoot to in order to solve the corners. It's intuitively pretty clear. Assuming you start with the buffer piece solved, There can at most be 3 cycles you have to break into. Two of these will be of length 2 and the other of length 3, and you will have to shoot to 3 stickers to solve a 2-cycle and 4 to solve a 3-cycle. 3+3+4=10
What about four 2-cycles?
And does it matter if the buffer is not in a cycle?

Anyway, I used GAP to get the distribution of cycle structures for permutations of 8 objects.
Code:
```Even Permutations
{ 7 }		5760
{ 6, 2 }	3360
{ 5 }		1344
{ 5, 3 }	2688
{ 4, 4 }	1260
{ 4, 2 }	2520
{ 3, 3 }	1120
{ 3, 2, 2 }	1680
{ 3 }		 112
{ 2, 2, 2, 2 }	 105
{ 2, 2 }	 210
{  }		   1

Odd Permutations
{ 8 }		5040
{ 6 }		3360
{ 5, 2 }	4032
{ 4, 3 }	3360
{ 4, 2, 2 }	1260
{ 4 }		 420
{ 3, 3, 2 }	1120
{ 3, 2 }	1120
{ 2, 2, 2 }	 420
{ 2 }		  28```

9. Awesome stats! Thanks dude.

So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle. The rest do. I hate when that happens. I have to go back and count all the corners by putting my fingers on them.

if my maths are correct, according to his signature, Amostay2008 "pwns" on 3360/40320 or 8.3% of corner solves.

10. Originally Posted by toastman
So, if my maths are correct, 16065/40320 or 39.8% of corner solves do *not* require breaking into a new cycle.
To get 16065, I see that you added up all the cases where there is at most one cycle. I note that if you use a fixed buffer, that buffer may not always be part of a cycle. For example, take the 7-cycle case. There are 5760 such permutations. 7/8 of them or 5040 will have the buffer in the cycle, and in the remaining 720 cases, the buffer piece is in its correct place (not part of the cycle). If you consider it to be "breaking into a new cycle" in this case when the buffer is not in a cycle to start with, but you still have a cycle to solve, then you have overcounted the number of cases of "not breaking into a new cycle."

Under these assumptions, the number of cases I calculate reduces to 13700 cases, or just a little under 34%.

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